Question

How do you find the roots, real and imaginary, of `y= 5x^2 - 6x - (2x+1)^2` using the quadratic formula?

Answer 1

Two Real roots at `x=5+-sqrt(26)`

Explanation:

`y=5x^2-6x-(2x+1)^2`

`y=5x^2-6x-(4x^2+4x+1)`

`rarr y=color(red)1x^2color(blue)(-10)xcolor(green)(-1)`

Applying the quadratic formula for roots:
`color(white)("XXX")x=(-color(blue)b+-sqrt(color(blue)b^2-4color(red)acolor(green)c))/(2color(red)a)`
we have
`color(white)("XXX")x=(-color(blue)((-10))+-sqrt(color(blue)((-10))^2-4xxcolor(red)1xxcolor(green)((-1))))/(2xxcolor(red)1)`

`color(white)("XXX")x=(10+-sqrt(100+4))/2`

`color(white)("XXX")x=5+-sqrt(104)/2`

`color(white)("XXX")x=5+-sqrt(26)`